Question

${{10}^{-6}}$ M NaOH is diluted by 100 times. The pH of diluted base is?
(A)- Between 6 & 7
(B)- Between 10 & 11
(C)- Between 7 & 8
(D)- Between 5 & 6

Answer 1

pH of a solution is calculated from the concentration of ${{H}^{+}}$ ions in the solution and is given as $pH=-\log \left[ {{H}^{+}} \right]$.
Similarly, pOH measures $O{{H}^{-}}$ ion concentration in a solution, i.e. $pOH=-\log \left[ O{{H}^{-}} \right]$.
Relationship between pH and pOH, based on the equilibrium concentrations of ${{H}^{+}}$ and $O{{H}^{-}}$ is
$pH+pOH=14$

Complete answer:
Let us solve the given question step by step.
Given, molar concentration of NaOH base = ${{10}^{-6}}$ M
NaOH being a strong base dissociates completely in water into $N{{a}^{+}}$ and $O{{H}^{-}}$ ions
$NaOH(aq)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)$
Since, one mole of NaOH is equal to one of $N{{a}^{+}}$ and $O{{H}^{-}}$. Thus, we can draw the following conclusion
$\left[ NaOH \right]=\left[ N{{a}^{+}} \right]=\left[ O{{H}^{-}} \right]={{10}^{-6}}M$
It is given that the base has been diluted by 100 times. So, the concentration of $\left[ O{{H}^{-}} \right]$now becomes ${{10}^{-8}}M$, i.e.
$\dfrac{{{10}^{-6}}M}{100}={{10}^{-8}}M$
To find the total concentration of $O{{H}^{-}}$ ions, i.e. $\left[ O{{H}^{-}} \right]={{10}^{-8}}M+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$, we need to consider the concentration of $O{{H}^{-}}$ from water.
We know that water ionizes as
${{H}_{2}}O\rightleftarrows {{H}^{+}}+O{{H}^{-}}$
One mole of water gives one mole of ${{H}^{+}}$ and $O{{H}^{-}}$,thus, we have
$\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]$
Ionic product of water, which is the product of concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ ions, at 25$^{o}C$ is ${{10}^{-14}}$. Since $\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}$, we can write that the concentration of $O{{H}^{-}}$ from ionization of water, ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}={{10}^{-7}}M$.
Therefore, total $\left[ O{{H}^{-}} \right]$ ions after substituting the value of ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$ will be
$\begin{aligned} & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}} \\\ & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}} \\\ \end{aligned}$
Multiplying and diving ${{10}^{-7}}$ by 10 in the above equation, we get
$\begin{aligned} & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}}\times \frac{10}{10} \\\ & \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-8}}\times 10 \\\ \end{aligned}$
Taking ${{10}^{-8}}$ common in the equation for simplification, we obtain
$\begin{aligned} & \left[ O{{H}^{-}} \right]={{10}^{-8}}(1+10) \\\ & \left[ O{{H}^{-}} \right]=11\times {{10}^{-8}} \\\ \end{aligned}$
Now we have the total concentration of $O{{H}^{-}}$, i.e. $\left[ O{{H}^{-}} \right]=11\times {{10}^{-8}}M$, we can find pOH as
$\begin{aligned} & pOH=-\log \left[ O{{H}^{-}} \right] \\\ & pOH=-\log \left[ 11\times {{10}^{-8}} \right] \\\ \end{aligned}$
Applying $\log (mn)=\log m+\log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as
$\begin{aligned} & pOH=-(\log 11+\log {{10}^{-8}}) \\\ & pOH=-(\log 11-8\log 10) \\\ \end{aligned}$
We know that ${{\log }_{10}}10=1$, on substituting it, the above equation becomes
$\begin{aligned} & pOH=-(\log 11-8) \\\ & pOH=-1.0414+8 \\\ & pOH=6.9586\approx 6.96 \\\ \end{aligned}$
To find the pH from pOH, we have the relation that is true for solution at 25$^{o}C$. Putting the value of pOH = 6.96, we have the pH of the solution
$\begin{aligned} & pH+pOH=14 \\\ & pH=14-pOH \\\ & pH=14-6.96 \\\ & pH=7.04 \\\ \end{aligned}$
Therefore, the pH of diluted base is 7.04, which lies within 7 to 8.
So, the correct answer is “Option C”.

Note: Note that concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ is important for dilute solutions. We cannot ignore the concentration of $O{{H}^{-}}$ due to water in this case, as the solution has become very dilute due to the addition of water. Due to the decrease in the number of $O{{H}^{-}}$ (and ${{H}^{+}}$) ions per unit volume, the pH of the basic solution has been reduced.